\(\left|x-\dfrac{3}{5}\right|< \dfrac{1}{3}\Leftrightarrow\dfrac{-1}{3}< \left(x-\dfrac{3}{5}\right)< \dfrac{1}{3}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{3}{5}>\dfrac{-1}{3}\\x-\dfrac{3}{5}< \dfrac{1}{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{-1}{3}+\dfrac{3}{5}\\x< \dfrac{1}{3}+\dfrac{3}{5}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{4}{15}\\x< \dfrac{14}{15}\end{matrix}\right.\)
vậy \(\dfrac{4}{15}< x< \dfrac{14}{15}\)
Vì giá trị tuyệt đối của \(x-\dfrac{3}{5}\) luôn > = 0
=> \(x\in\varnothing\)
\(•nếu\:x\ge0\:thì\:\left|x.\dfrac{3}{5}\right|=\dfrac{3x}{5}\\ •nếu\: x< 0\: thì\: \left|x.\dfrac{3}{5}\right|=-\dfrac{3x}{5}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3x}{5}< \dfrac{1}{3}\left(với\:x\ge0\right)\\-\dfrac{3x}{5}< \dfrac{1}{3}\left(với\:x< 0\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x< \dfrac{5}{9}\left(nhân\:0\le x< \dfrac{5}{9}\right)\\x>-\dfrac{5}{9}\left(nhận\:-\dfrac{5}{9}>x\ge0\right)\end{matrix}\right.\)
vậy pht có nghiệm là \(-\dfrac{5}{9}< x< \dfrac{5}{9}\)
